1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Partial Implicit Differentiation

  1. Jun 11, 2005 #1
    Just wondering if I did this right:

    Here is the question: find [tex]\frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1[/tex]

    Now I put the [tex] \frac{\partial z}{\partial x} [/tex] on both sides then got.

    [tex] \frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0 [/tex]


    [tex] \frac{\partial z}{\partial x} = -\frac{2x}{9z} [/tex]

    Now I know this is the right answer I am just curious if I did it right, it has been a while.

    Now my reasoning:

    [tex] \frac{\partial z}{\partial x} \frac{x^2}{9} [/tex] is just the derivative with respect to x, so it will be [tex]\frac{2x}{9}[/tex]

    [tex] \frac{\partial z}{\partial x} \frac{y^2}{4} [/tex] has no z or x, so it is constand and therefore 0.

    [tex] \frac{\partial z}{\partial x} \frac{z^2}{2} [/tex] has a z, so it is the derivative of itself, but times [tex] \frac{\partial z}{\partial x} [/tex]

    Is all of that correct, or did I do something wrong. The book I have only shows one example :mad:

    Last edited: Jun 11, 2005
  2. jcsd
  3. Jun 11, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    It looks okay.You could have done it using the theorem of implicit functions,as well.

  4. Jun 11, 2005 #3
    Thanks dex! I have no idea what the theorem of implicit functions is, but I probably could not use it anyway.
  5. Jun 11, 2005 #4
    One more question for now :smile:

    I need to find

    [tex] \frac{\partial z}{\partial x} sin(xz) [/tex]

    How exactly does this work?

    Will it be: cos(xz) [times] what? [tex] \frac{\partial_}{\partial x} xz [/tex]

    Which will then be cos(xz) [times] x ? This one seems to be wrong. Any ideas?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook